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Exam 2 Study Guide

by: Gerry Otto

Exam 2 Study Guide MATH1045

Gerry Otto
GPA 3.5
Applied Calculus 2
Ryan Therkelsen

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Applied Calculus 2
Ryan Therkelsen
Study Guide
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Popular in Mathematics (M)

This 12 page Study Guide was uploaded by Gerry Otto on Wednesday October 21, 2015. The Study Guide belongs to MATH1045 at University of Cincinnati taught by Ryan Therkelsen in Fall 2015. Since its upload, it has received 23 views. For similar materials see Applied Calculus 2 in Mathematics (M) at University of Cincinnati.


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Date Created: 10/21/15
61 62 Math 1045 Exam Study Guide Oct 23rd 530630 6167 Antiderivative An antiderivative of a function f is a function whose derivative is f In other words F is an antiderivative off if F39 f To nd an antiderivative for a function f we can often reverse the process of differentiation The Inde nite Integral We introduce a notation for the family of antiderivatives that looks like a de nite integral without the limits We can If H m the inde nite integral 01 f if if all antiderivatives ff 151 F 1 of f if are of the form F iJi39i 395 we write 0 It is important to understand the difference between f and f 0 Example Find far m Since H ifi 3134 is one antiderivative of f ifi all other antiderivatives of f lfl are of the form H if 39539 Thus fax3 it EJ394 if The function FX is called an antiderivative of ix if E ij flile The family of all antiderivatives of a function ix is called the inde nite integral of the function iX and is denoted by cfx Thus if F is a particular antiderivative we may write IfxjcfxFxjC De nite vs Inde nite Integral 0 With an inde nite integral there are no upper and lower limits on the integral here and what we39ll get is an answer that still has x39s in it and will also have a K pus K in it o A de nite integral has upper and lower limits on the integrals and it39s called de nite because at the end of the problem we have a number it is a de nite answer If Est 9d 0 Example Evaluate the following inde nite integeral 0 We notice it is asking for the most generalantiderivative so we recall that If F iii is any anti derivative of f Ii then the most general anti derivative of f Ii is called an indefinite integral and denoted Iffci1f FIJI r is an ennstsnt J squot 3I Ei39nl xz g gx 9I 39 0 So 63 Fundamental Thenrem nf Calm ue f is continuous en the interval m is and F an antiderivatieenf lhstisif F s f f in em s bi s sin I Example 1 39339 K Eninpnte Jr 2 iiiquot ninneneslly and using Ellie Fundamental Thenrern 1 Slutin Using a eslenlntnr we nbtein f 2 iiiquot 1 The Fundamental Tlienrein alien395 us in enmpute the integral exactly An entitlerii39stis39e ni39 ti 21quot is F ti t39 and we nbtsin fEJquot n39JFl3i Fllj 32 13 1 tmprper Integrate a Statistri in taint disentaintt if the definite integral f r tit Ewe have assumed that the internal t39 g 1 g is is and I1 finite length and the integrand is E fI39IIiITIIJJDIJJEi Empr per integrate is a definite it39li gf i in 1athieh tine er hath tilquot iimi ts tad iti tegra titttt is in nite er the inte A is llllh ll tji di dirt estatittpiae tifattt impr p r ilfll gf l is In 1 tit 1 3 o Convergent Improper Integrals i Cnnsidet a tunetinn x athleh exhibits a Type i at Type I hehatier en the interaai an in ether wards the integral 1 f in is impmpert We saw hetnre that the this integrai is de ned as a limit Theretnre we hate El twe eases i the limit exists and is a nnnthert in this ease we say that the implmner integtai is enlistment 2 the limit dees not exist at it is in nite then we satr that the intranet integral is divergent t it the internner integtai is split late a sum nf intnteper integrals heeanse t presents mete than nne innnrepethehatiet an in til then the integral ennterges if and only it antr single ltt ipt dfhttl integral is I C h g it Example 4 Interpret and estimate the value of Solution This integral represents the area under the graph of from 1quot 1 in nitely far to the right See Figure 120 t39 F39igure 52a Area representation of improper integral To estimate the value of the integral we find the area under the curve between t l and t l where b is large The larger the value of b the better the estimate For example for b if l 1009 101 V 391001 illDO 1 M f sitquot 09 f sit 199 f cit H999 1 t39 1 t39 1 t39 These calculations suggest that as the upper limit of integration tends to infinity the estimates tend to III We sar that the improper integral f cit converges to To show that integral converges to exactly I and not to LEIUCII sari we need to use the Fundamental Theorem of Calculus See Problem 29 It may seem surprising that the region 1 t39 shaded in Figure 121 which has in nite length can hauefmite area The area is finite because the values of the function 1 l39 Aquot shrink to zero very fast as 1quot gt ea In other examples where integrand does not shrink to zero so fast the area represented by an improper integral map not be finite In that case we say the improper integral diverges See Problem 43 64 So pply and Demand Curves As we saw in Chapter I the quantity of a certain item produced and sold can be described by the supply and demand curves of the item The supply curve shows what quantity q of the item the producers supply at different prices p The consumers beiiasior is reflected in demand curve which shows what quantityr of goods are bought at 1serious prices See Figure 121 p prieslunill ill Swat Demand g qu entity Figure 521 Supply and demand curves t is assumed that market settles at the equilibrium price pE and equilibrium gummy at where the graphs cross At equilibrium 1 quantity aE ofan item is produced and sold for a price of pE each 0 Producer Surplus Producer surplus or producers39 surplus is the amount that producers bene t by selling at a market price that is higher than the least that they would be willing to sell for this is roughly equal to pro t 0 Consumer Surplus Consumer surplus is the difference between the total amount that consumers are willing and able to pay for a good or service indicated by the demand curve and the total amount that they actually do pay 0 Equilibrium Price The price at which the quantity of a product offered is equal to the quantity of product in demand r r 4 n r C onre Limer surplus at price p Area bemreen demmici e UWE and horizontal line or p 7 r I 1 Proclue er surplus or price p p Sr39uniiti QM Consumer surplus p 1 8f 39 il i 4 2 p I Sniun il u x i i 11 Producer surplus p 77 817 g quot5 geumu39wumm o Example i Q Eiiiiiji d r r rr rrlliirr lrrr 4 Are premieren suppi cuwe Mid horizontal lime or p U D u ijuuitsi U1 D q U iiS The euppigyr and demand curves for a product are given in Figure 625 a What the equilibrium price arui quantity p IISI39Uniil 21g If 2 E i I D 39 Munvanity q39 some leuooo Figure 625 Supply and demand curves for a product in Ar tire equiiibrium price oalo uiete and interpret the consumer end producer surplus a The equilibrium prise is pi l 3039 unit and the equilibrium quantin ql 30000 units h The eensumer surplus the area under the demand eurire abuse the TIE p 30 See Figure 020 We have Censumer surplus Ares pf tritu igle lE use Height S 163 Eli tiedCELEBS This tells us that sunsumers gsiu 04IH300 in buying at the equilibrium prise instead of at the price they wnu ld have been willng to pay 339 Sin nit 244 i gimsiil l39iliEr EUI Tll LUEv pquot Fifi i if 2 80000 90000 Figure 626 unsurner surplus i units Predueer surplus Ares pf triangle 113 nse Height 2 S 34 3200000 Sn producers gain 3200000 by supplying grinds at the equilibrium priee instead of the prise st whieh wnuid have been willing to provide the goods 65 5 it delleu shreeu p iquotunii quot10 Predueer surplus LIquot 3ilJl39UllSl is 50pm 120000 Figure 521 Predueer surplus Income Stream 0 When we consider payments made to or by an individual we usually think of discrete payments that is payments made at speci c moments in time However we may think of payments made by a company as being continuous The revenues earned by a huge corporation for example come in essentially all the time and therefore they can be represented by a continuous income stream Since the rate at which revenue is earned may vary from time to time the income stream is described by 0 Notice that 3 iii is a rate at which payments are made its units are dollars per year for example and that the rate depends on the time 3 usually measured in years from the present 0 Present and Future Values of an Income Stream 0 Just as we can nd the present and future values of a single payment so we can nd the present and future values of a stream of payments As before the future value represents the total amount of money that you would have if you deposited an income stream into a bank account as you receive it and let it earn interest until that future date The present value represents the amount of money you would have to deposit today in an interestbearing bank account in order to match what you would get from the income stream by that future date 0 When we are working with a continuous income stream we will assume that interest is compounded continuously If the interest rate is rquot the present value P of a deposit 5 made it years in the future is Pawt 1esexnt value f if r i e 39 air f U ll 39 i quot ii Future is alu e 2 Fire sent v alue e 0 Example 0 Q Find the present and future values of a constant income stream of 1000 per year over a period of 20 years assuming an interest rate of 6 compounded continuously Selutiun Using 539 If t l 2 and r GLUE and a ealeulatet er eetnputer tun evaluate the integral we have a gt V viii t39 F 39 39 Prawn Mme f 1003 a an r Eli l j K w rm a tn v r u 5 f39l r Alternatee1 aging the fundamental lheeretn and the fact that an antidenaatiae at 110006 3393quot 1a m f we have if quotr r a W m in We can get the future value B from the present value P using Fe r ae Future value l 154539QE 1m 2 0 Tip On PV and FV problems switching the time units either by calling for quarterly or monthly compounding or by expressing time in months and the interest rate in years is an oftenused tactic to trip up test takers who are trying to go too fast Remember to make sure the units agree for r and N and are consistent with the frequency of compounding prior to solving 66 o In this section we introduce integration by substitution which reverses the chain rule According to the chain rule lnaltle 5 quot39 3 Exll39 7 3 il tl Eaten l tatau tat Haretithe at amide quot Eleai mljj ue imuaa 0 Thus any function which is the result of differentiating with the chain rule is the product of two factors the quotderivative of the outsidequot and the derivative of the insidequot If a function has this form its antiderivative is flat3513 739 0 Exa m ple Use the chain rule to find and then write the corresponding antidifferentiation formula d a r a h f lJfl rr3 if c f a 1HEIJ43 3 Using the chain rule we see i 39I I E I It all if i a so e a b Using the chain rule we see 6 iii 1 u 9 39i t I u39 I 39i Il Iquot 39I S i ll a Ill 0 7i Illiquot ll ll Ilii ill so li il trig 39iquot ll 0 it quota b quotn quot J 39 39 b i C Using the chain rule we see l in 2 4 if It so f a 17 311 it in I C 7 J 39 39 nquot 4 iquot 4 39 o In Example 1 the derivative of each inside function is 31 Notice that the derivative of the inside function is a factor in the integrand in each antidifferentiation formula 0 Finding an inside function whose derivative appears as a factor is key to the method of substitution We formalize this method as follows To Make a Substitution in an Integral nil Let n be the inside function and illl 11quot If i it nit Then express the integrand in terms of 11 0 Example Make a substitution to find each of the following integrals a cili 21 at h f 19 1 21 flit c f 1 gig m t 4 3953 We i k fer an inside fittietieti whewe derivative appears Eh feetet Ilt thie CHE 5 the initiate ttietieti ie 1 39 l with derivative 21 We iet H II tt39ttt w I fitquot 2 fitquot gim l integrand eat new he rewritten in terrmt ttf w Fl f eff21 fitquot f equotquot39 etw equotquot39 C elf 39339 Eh Here ineiuie fwttetieti i5 1 2 1 i with derivative 21 t We let it J 2 1 Theti at39w w ij at 21 titquot t Rewriting the erighte integral in tetme ef wt we have f 511 eta thw 5 tt39w Elw if 31 t 11E39 if Again 3 ehttltgihg the h39 fi ijl te it 1 we Simplified integrative E 1 The il39tElEiE futietleti 15 JE39 4 en we let it 1quot 4 5th w I ti 21 titquot a Swhetututmgi we have 67 New ititrtttitiee i39nfegmfiirm by Farm 1whitrh 15 a teehtiique hf nding integrals based mt the prettttet miet We begin with pmehtet title I I l 7 quot 5 5 tit H1 tt1 l l where it and 15 are tttetjntie ef 1 with derivatives it and 15 iteetitreljltt in t39e rewrite this he I l viv55 tit ii hi hi and then integrate hath sides id5 1n i gAn d 51n fin tit fat tit tit fut tit Intwg ratiuri by Pa l 39 fin tiltquot H155 fit1 t ilf 0 Example UR intagr tin In part5 In find f 1 63quot ails Enluti n l Wt 1amp1 3 at ch r Elf m 5311 chime H J and 13 1 1111153 54 l and 13 E39Ti 7 gee1391 513 m gee1391 7 11 e1391 aixJEI EIC 39 i545 Haw tn Ehuuse H and If I l Wh l f f 3m 1m 15 has 3mm r d m be h m nd 15 g l l I I1 h lp if in 15 L5 impkf Hf E11 l 39l um mare manplimt than m E 0 Example 2 Find f Jc ln 1 ail Enlutinn We can write Jc ln 1 35 41 where a 1n 1 and 15 Jc ThEtr1 15 l 1quot and a l in 5n integrmi n by part5 givga ME f f l 1 11 In ch IE39 11 111ch inf x 11 I111I IIEI 11 fligxiln I J C TI Examp eda WE did watchman 1 1n 1 b it is 11m imm n i t y EIEEELF what 15 wnxu d In 122ml5 WE i mgmtjnm y 3115 in Examplc 3 m nd the amiderl ymiw If in 1 Maui using it 1n 1 5 35 we haw aunts giws 4 15 Jc I T 1 which i5 simpler m ima gmtc than 41 Jc 1n 1 mmp Shaw that it aims ml haw In first fiacl r in ha integrand ham IE39ZIL


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